LeetCode Copy List with Random Pointer

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A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

思路分析：这题要求拷贝链表，包括内容，next指针和random指针。容易想到的思路是定义一个hashmap保存旧链表节点到新链表节点的映射关系，第一次遍历链表，只拷贝next指针和node，并且init hashmap，第二遍遍历链表再拷贝所有节点的random指针,借助hashmap可以在O（1）时间内由给定旧链表节点找到新链表对应节点。时间复杂度和空间复杂度都是O（n）。

另有一个解法不需要额外空间，需要三遍遍历，要仔细画图分析，详细可见

http://codeganker.blogspot.com/2014/03/copy-list-with-random-pointer-leetcode.html

http://blog.csdn.net/fightforyourdream/article/details/16879561

AC Code

/** * Definition for singly-linked list with a random pointer. * class RandomListNode { *     int label; *     RandomListNode next, random; *     RandomListNode(int x) { this.label = x; } * }; */public class Solution {    public RandomListNode copyRandomList(RandomListNode head) {        //0133        if(head == null) return null;        RandomListNode newHead = new RandomListNode(head.label);        HashMap<RandomListNode, RandomListNode> oldToNewMap = new HashMap<RandomListNode, RandomListNode>();        oldToNewMap.put(head, newHead);        RandomListNode newCur = newHead; // iter for newList        RandomListNode cur = head.next; // iter for oriList                while(cur != null){            RandomListNode newNode = new RandomListNode(cur.label);            newCur.next = newNode;            oldToNewMap.put(cur, newNode);            newCur = newNode;            cur = cur.next;        }                newCur = newHead;        cur = head;        while(cur != null){            newCur.random = oldToNewMap.get(cur.random);            cur = cur.next;            newCur = newCur.next;        }        return newHead;    }    //0202}

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